Covariance properties 2

Probability Level 2

Given knowledge of $\text{Cov}(W, Y)$ , $\text{Cov}(W, Z)$ , $\text{Cov}(X, Y)$ , and $\text{Cov}(X, Z)$ , which of the following can necessarily be computed?

I. $\text{Cov}(W + X, Y + Z)$

II. $\text{Cov}(Y + Z, W + X)$

III. $\text{Cov}(W, X + Y + Z)$

IV. $\text{Cov}(W, X + Y + Z)$ , if it known that $W$ and $X$ are independent

I, II, III, and IV I, III, and IV only I only I and II only I, II, and IV only

1 solution

Arsalan Khan
Jul 25, 2018

expression III. cannot be computed because we have no information regarding Cov(W,X). For expression IV. knowing that W and X are independent makes Cov(W,X) = 0. Thus making IV. computable.

Interesting... It looks to me that we have information with regards to how Y changes with respect to a change in W and how X changes with respect to a change in Y thus connecting the 2 together we get that when we change W => the Y changes and given that we know how the change in Y affects the change in X we will also know how X changes.

Stepan Šklarik - 3 months, 3 weeks ago

Covariance properties 2

1 solution

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